TITLE: Field Norm Surjective for Finite Extensions of $\mathbb{F}_{p^k}$
QUESTION [8 upvotes]: I'd rather not have the answer, because I feel like this should be a relatively easy question, and I'm just missing some key step, but could anyone give me a hint on showing that the norm (defined as $N(a)=\det(L_a)$) where $L_a$ is the linear transformation given by multiplication by $a$ is surjective in the case of a finite extension of a finite field?
I've been looking at the fact that $N(x)=x^n$ for any $x$ in the base field, where $n$ is the degree of the field extension.  But this doesn't necessarily give me back every element in the base field, for instance $\mathbb{F}_3(\sqrt{2})$, where we have to apply norm to $\sqrt{2}$ to get $2$ back.  Is there a basic fact about finite fields maybe that I'm missing, or something more clever regarding a field extension?
Thanks!

REPLY [7 votes]: Remark: I replied too soon and provided an answer for the trace of a field extension instead. 
I won't delete it because it's somewhat interesting, even though it's useless.
I have included but different calculation for the norm just for completion of my reply.
$$ N\colon \mathbb F_{q^\ell}^*\to \mathbb F_{q^\ell}^* \colon x\mapsto xx^q x^{q^2}\cdots x^{q^{\ell-1}}  = x^{\tfrac{q^\ell-1}{q-1}} $$
Notice that $N$ is a morphism of cyclic groups and that in fact $\mathrm{im} N \subseteq \mathbb F_q^*$. 
Now it's well known (or else easy to show) that if 
$$ \varphi \colon C\to C \colon x\mapsto x^m $$
is a morphism of cyclic groups, then $|\mathrm{im} \varphi| = |C|/(m,|C|)$. 
In this case $|\mathrm{im} N |= \frac{q^\ell-1}{(q^\ell-1,\frac{q^\ell-1}{q-1})} = q-1$ and indeed $N$ is onto $\mathbb F_q^*$.

Original reply:
I'm not sure where exactly you are stuck, so I wouldn't know what to hint, but here is how I usually deal with such a problem.
Let $\mathbb F_{q^\ell}/\mathbb F_q$ be a field extension, with associated trace
$$ T\colon \mathbb F_{q^\ell}\to \mathbb F_q \colon x\mapsto x+x^q + x^{q^2} + \dots + x^{q^{\ell-1}} $$
Notice that $T$ is a morphism of additive groups. (And note that indeed $T(x)^q = T(x)$ for all $x$ so it goes to $\mathbb F_q$ indeed).
Now first of all, all elements in $\ker T$ are solutions to an equation of degree $T^{\ell-1}$ therefore $|\ker T|\le q^{\ell-1}$. On the other hand $|\mathrm{im}\, N| = q^\ell/ |\ker N| \leq q$ therefore $|\ker T| \geq q^{\ell-1}$. Therefore $|\ker T|=q^{\ell-1}$ and $|\mathrm{im} T|=q$ and $T$ is a surjection.
-- Edit. I should clarify that the definition of "Trace" I use may seem different from yours; fortunately Arturo Magidin --as always-- provides an detailed explanation on that.